Question

The specific heat of a substance at temperature $t^{\circ }C$ is $s=a{t}^2+bt+c$. The amount of heat required to raise the temperature of ${}^{\prime }{m}^{\prime }$ grams of a substance from $0^{\circ }C$ to $t^{\circ }C$ is

• A. $m(\frac{t_0^{3}a}{3}+\frac{b{t}_0^2}{2}+c{t}_0)$
• B. $m(a{t}_0^3+b{t}_0^2+c{t}_0)$
• C. $m\left(\frac{t_0^{3}a+b{t}_0^2+c{t}_0}{2}\right)$
• D. $m(\frac{t_0^{3}a}{2}+\frac{b{t}_0^2}{3}+\frac{c{t}_0}{2})$

Answer 1

The correct option is A $m(\frac{t_0^{3}a}{3}+\frac{b{t}_0^2}{2}+c{t}_0)$

Amount of heat required $H={\int }_0^{t_o}msdt$
Or $H={\int }_0^{t_o}m(a{t}^2+bt+c)dt$
Or $H=m\times$ $(\frac{a}{3}{t}^3+\frac{b{t}^2}{2}+ct){|}_0^{t_o}$
Or $H=m\times$ $\left[\frac{a}{3}\right(t_o^3-0)+\frac{b(t_o^2-0)}{2}+c(t_o-0\left)\right]$
$⟹H=m(\frac{a{t}_o^3}{3}+\frac{b{t}_o^2}{2}+c{t}_o)$